I teach a precalculus course almost every semester, and over these semesters I've found various things that work quite well. For example, when talking about polynomials and rational functions, in particular "zeroes" and "vertical asymptotes", I introduce them as the same thing, only the asymptotes are "points at infinity". This (projective plane) model helps the students in understanding what multiplicity of a zero/asymptote really means. I later use the same projective plane model to show how all conic sections (circle, ellipse, parabola, hyperbola, lines) are related. I get very nice feedback on this, and the students seem to really enjoy it.

I have not, however, been able to find similar motivating examples for introducing complex numbers. I know there must be similar (pictorial!) arguments to engage the students and pique their curiosity, but I haven't found it yet. Simply saying "all polynomials have a zero over the complex numbers" doesn't really do it for them (again, the more pictures involved, the better).

Are there "neat" and "cool" ways of talking about complex numbers for the first time, that are understandable by beginning precalculus students, but also interesting enough to capture their attention and provoke thought?

One. Use Euler's formula for conversions between polar and rectangular coordinates in 2 dimentions. You can also use it for rotations. This method needs less handwriting and thiking than using sin, Cos, and tan. Notation is more compact.

$$e^t=\cos(t)+i\sin(t)$$

$t$ is angle in radians. This goes with the vector approach. Multiply above equation by $R$ and then you get $(R,t)$ are polar coordinates on the left side. Then on the RHS the real and imaginary parts are the rectangular coordinates.

A rotation is evident by multiplying by e^(if) or adding if in the exponent, where f is the rotation angle, phi, in radians. (2π is the full distance around a circle of radius 1.)

Google or hit the library. It might even be in your book.

Modern scientific calculators and math software handle complex numbers seamlessly. This makes use even easier. Ie if you entered sqare root of -1 it would give you 0+i or (0,1) in complex retangular coordinares. Trig, exponentioal, and log functions for sure. You can ask your class to test their calculators to find the good ones.

This shows relations between e, trig fuctions, ln, pathagorean therom, and π. Trig functions can be derived from the equation. Talor series expantions of the equation also work.

Note all functions above work with complex inputs and can provide complex outputs.

i know i am going to get a lot flack for this but why would take time off in pre cal to introduce complex numbers? the whole point of pre cal is to prepare the students for a two semester long calculus sequence. you hardly use any complex numbers at all in the two semesters of calculus. is not the time better spent getting them ready to do well in calculus.

Edan Maor provides an answer on this post, which gives a nice explanation:

A way to solve polynomials

We came up with equations like $x - 5 = 0$, what is $x$?, and the naturals solved them (easily). Then we asked, "wait, what about $x + 5 = 0$?" So we invented negative numbers. Then we asked "wait, what about $2x = 1$?" So we invented rational numbers. Then we asked "wait, what about $x^2 = 2$?" so we invented irrational numbers.

Finally, we asked, "wait, what about $x^2 = -1$?" This is the only question that was left, so we decided to invent the complex numbers, in particular "imaginary" numbers, to solve it. All the other numbers, at some point, didn't exist and didn't seem "real", but now they're fine. Now that we have complex numbers, we can solve every polynomial, so it makes sense that that's the last place to stop.

Pairs of numbers

This explanation goes the route of redefinition. Tell the listener to forget everything he knows about imaginary numbers. You're defining a new number system, only now there are always pairs of numbers. Why? For fun. Then go through explaining how addition/multiplication work. Try and find a good "realistic" use of pairs of numbers (many exist).

Then, show that in this system, $(0,1) * (0,1) = (-1,0)$, in other words, we've defined a new system, under which it makes sense to say that $\sqrt{-1} = i$, when $i=(0,1)$. And that's really all there is to imaginary numbers: a definition of a new number system, which makes sense to use in most places. And under that system, there is an answer to $\sqrt{-1}$.

Explain the history of the imaginary numbers. Showing that mathematicians also fought against them for a long time helps people understand the mathematical process, i.e., that it's all definitions in the end.

I'm a little rusty, but I think there were certain equations that kept having parts of them which used $\sqrt{-1}$, and the mathematicians kept throwing out the equations since there is no such thing.

Then, one mathematician decided to just "roll with it", and kept working, and found out that all those square roots cancelled each other out.

Amazingly, the answer that was left was the correct answer (he was working on finding roots of polynomials, I think). Which lead him to think that there was a valid reason to use $\sqrt{-1}$, even if it took a long time to understand it.

I like the (macro-)historical perspective in the section "A way to solve polynomials" :-D
–
user1551Dec 5 '12 at 17:44

5

It was Bombelli, showing that for cubics with three distinct real roots, the Cardano Formula involved square roots of negative numbers. It turns out that one can solve this "real" problem algebraically only by travelling through the complex numbers.
–
André NicolasDec 5 '12 at 18:32

2

I would say calling them "imaginary" numbers is evil. It throws off the learner gives them an impression that we are talking about some fairytale world and these numbers have nothing to do with the actual world. Why would some numbers be imiganary? They are as "real" as R, and are used to solve very real problems
–
MidhatDec 6 '12 at 8:19

15

I don't see the point of copying this answer when it could just be linked. Regardless, attribution should be given for copied work.
–
Jonas MeyerFeb 21 '13 at 17:47

6

I decided to try Googling the only part of this answer that hadn't been attributed to another source in the updated version. It led me to this MathOverflow answer.
–
Jonas MeyerFeb 22 '13 at 4:23

This is like Javier and Izkata's answers combined. Go to this site that plots complex function graphs and type in $$f(z)=z^2+1$$ then look at the top view. You will see the solutions are not along the horizontal center line. You could produce several of these graphs and ask students in groups to find out how they relate to the roots of the function.

Note: the following is commonly known and you can find several websites that give the following example.

I have taught complex numbers several times and I do agree with you that it can be hard to motivate. While I agree that pictures often can be helpful in motivating and understanding a problems, I also don't think that one has to be afraid of staying abstract.

I have done this with some success in the past: First recall how we have the formula that solves a quadratic equation. And note how nice it is, that one doesn't even have to think, one just has to remember a formula. Now then mention that historically there was an interest in solving the general cubic equation. Without (or maybe you would want to do this) actually showing the general formula mention that at some point people considered this special cubic equation (in fact all cubics can be brought into this form, but if we only allow positive coefficients (since they are nicer) consider just the following):
$$
x^{3} = px + q.
$$
where $p$ and $q$ are positive real numbers. Now the way this was solved was by letting $x = u + v$. Then
$$x^3 + px = u^3 + v^3 + 3uv(u + v) − p(u + v) = q.$$
And so
$$
\begin{align}
u^3 + v^3 &= q \\
u^3v^3 &= \left(\frac{p}{3}\right)^3.
\end{align}
$$
So to solve the cubic, one could try to find $u$ and $v$ because $x = u+v$, that is: Find two numbers whose sum is $q$ and whose product is $(p/3)^3$. That sounds like a simple problem and the solution is (think about this for a moment)
$$
\begin{align}
u &= \sqrt[3]{\frac{1}{2}q + w} \\
v &= \sqrt[3]{\frac{1}{2}q - w}
\end{align}
$$
where
$$
w = \sqrt{\left(\frac{1}{2}q\right)^2 - \left(\frac{1}{3}p\right)^3}.
$$
Now try to apply this to the equation:
$$
x^3 = 15x + 4
$$
Then $w = \sqrt{-121}$. Opps. I guess you have to stop now since we can't take a square-root of a negative number. Hmm... maybe if we just continue ignoring that for a moment (the rebels that we are) we see that we actually get
$$
x = u + v = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}
$$
Magic happens and we actually get
$$
x = 4.
$$
So the point is (IMO) that ignoring that we can't really take a square root of a negative number lets us find a real solution to a cubic equation. Of course complex numbers have a life outside being used to find real zeros, but at least it shows how some people started thinking about them. Then you can say that now you are going to make it formal what complex numbers are and then continue from there.

Obviously this way of introducing complex numbers will take some time and there is some "hard" algebra involved, but usually I get the feeling that it does get the point across. I like the method because the final example is easy to understand. You can get the calculator involved (or you can do the algebra if you would like) in the last step where you find the value of $x$. There is some history also. Just the basic sub-problem of finding two numbers is $4$ and whose product is $125$ can be interesting in itself.

I was never satisfied with the "magic happens" bit because finding the cube root of $2+ \sqrt{-121}$ is algebraically as hard as solving the original equation. suddenly saying "but this is a cube root of this" is as magical as saying "but $4$ is a solution to the original equation !" so you are reducing hard equations to solving hard cube roots of numbers that aren't even supposed to exist.
–
mercioDec 5 '12 at 17:48

1

@mercio: I think that one can do this example as an introduction or one can do it after having covered complex numbers. In the first case one would indeed have to simplify certain things. So as an introduction one could maybe just show on a calculator how x indeed is $4$. A point being that one never actually sees a complex number. Then later after you have covered complex numbers, you can go back to that "magic moment" and show how this is done by hand. No matter what, hopefully the point gets across that one can abstractly just work with $\sqrt{-1}$ without worrying too much about what it is
–
ThomasDec 5 '12 at 17:51

(cont.) and it will actually work as the example above shows. So as an introduction this works well (IMO) because it shows how complex numbers can be used to solve problems that we can understand from just knowing about real numbers.
–
ThomasDec 5 '12 at 17:53

I don't have the rep for such a minor edit, but one of you equations is missing an exponent. u^ + v^3 => u^3 + v^3
–
xanDec 5 '12 at 20:48

Unfortunately all the "cool" stuff connected with complex numbers is probably over the students' heads right now.

It's nice that you can find the roots of any polynomial and all, but all the "real action" is elsewhere.

Here is something that may be easy to explain. Because complex numbers are two-dimensional, a function with a complex domain and range has a geometric interpretation: it maps some two-dimensional space to another. So it can represent a geometric mapping, which morphs one space to another. That has not only obvious applications in computer-based visualization, but in problem solving: e.g. translating some problem with boundaries that are some oddly-shaped space into one over a rectangle.

Here is another. Because complex numbers are two dimensional, a complex-valued function can map each point in a two-dimensional space to a vector. In other words, complex functions can represent 2D fields. Properties of field, such as "fluxless" and "irrotational" can be connected to certain properties of the underlying functions which are nicely expressed using complex numbers.

And of course complex numbers have applications in signal processing, because signals exhibit frequency and phase, which give rise to two dimensions. The Fourier Transform uses complex functions, allowing a representation of a signal in the time domain to be taken into the frequency domain, with phase information. The magnitude of the complex number at each frequency tells us the amplitude of the signal at that frequency, and the angle of the complex number with respect to the real axis (the "argument") gives us the phase.

At least in the US, I think most students learn complex numbers before they learn about "fluxless" or "irrotational" vector fields.
–
Jesse MadnickDec 6 '12 at 5:55

Yes, well I think I touch on that point. See above: all the "cool" stuff connected with complex numbers is probably over the students' heads right now. However, some of these topics could be introduced in a very superficial way as "applications of complex numbers".
–
KazDec 6 '12 at 21:59

I really liked the explanation given by the site BetterExplained. Basically, you think of multiplication of real numbers as a geometric transformation.

Take any real number, and represent it as an arrow on the real line, starting from $0$. If you multiply it by a positive number, you change the arrow's length. If you multiply it by $-1$, you make the arrow point the other way. And multiplication by negative numbers other than $-1$ is just a reflection together with a scaling.

Thinking about multiplication this way, if you start with $1$ and multiply twice by $-1$, you reflect twice and therefore you get back to where you started. This makes sense, since $(-1)^2 = 1$. Now, is there an operation that, applied twice to $1$, gives $-1$? It's clear that no real number will do. But what if you escape the real line for a bit and rotate $1$ twice by $90^\circ$? You get $-1$. So now it turns out that the reflections that negative numbers were responsible for were just rotations by $180^\circ$, but we couldn't tell because we were restricted to just a line.

And now we could say that this rotation by $90^\circ$ deserves a name, so let's call it $i$. Of course, a rotation by $90^\circ$ in the other direction will also do, so we just pick one of them and call it $i$, and the other will be $-i$, because they're mirror images of each other.

The last step is realizing that now we have the whole plane to work with. We can use what we know about vectors and say that any arrow starting at $0$ and ending somewhere in the plane can be represented as a linear combination of $1$ and $i$.

Thinking this way, De Moivre's formula becomes a definition, which makes it clear what complex number multiplication really is (absolute values are multiplied while angles are added), because that's how we started! We wanted $i$ to represent a rotation by $90^\circ$, not some abstract solution of $x^2 = -1$.

I love this explanation because it is so intuitive, but it always makes me think, hey, it was so easy to escape from the real number line into two dimensions, let's just use that trick again to extend it to three dimensions... afaik it does not work (although 4 or more dimensions are possible), but I've never seen an equally intuitive explanation of why. (But maybe that's a topic for another question.)
–
Alistair BuxtonDec 5 '12 at 20:27

@AlistairBuxton: Consider which of the following i represents: Half of a reflection along the real line, an axis in 2D space, or a direction of rotation in 2D space? Now, is there any "trick" that will similarly represent half of a complex conjugation, an axis in 3D space, and a direction of rotation in 3D space (recall that you need a scalar component here to get a magnitude of 1!)?
–
camccannDec 5 '12 at 22:02

Not a very mathematical answer (sorry), but if you're looking for something visually interesting to introduce complex numbers, M. C. Escher has got you covered. I won't try to explain it, because I'll do a bad job. I'll let the real mathematicians do that:

The problem with this approach (which is how I do it now) is that it is indeed artificial, and students who mostly will not do much more math in their adult lives rarely find time to care about imaginary numbers! There are ways around this (as an example, drawing nice graphs of the parabola $x^2+1$, and showing where its "zeroes" went), but I don't know how to do that in a convincing and interesting way!
–
user641Dec 5 '12 at 17:07

@user1551: I am talking here of the viewpoint of "complexifying" the curve. This is something I have already thought about introducing (it explains why hyperbolas appear disconnected, while circles do not), but I find it highly nontrivial to do this in an elementary and convincing way.
–
user641Dec 5 '12 at 18:04

One of my favorite introductions to the complex numbers is to not define them with respect to the square root of $-1$! In fact, introduce the complex numbers and the operations therein as just an extension of those of the reals that they all know and love. Then, for instance, define addition and how this works in the plane (as vectors). Then, define multiplication, albeit in the necessarily "strange" way, saying that we need to be able to multiply these 2-tuples. Then proceed to blow their mind that actually $(0,1)$ is the square root of $(-1,0)$! It seems that the typical introduction to the complex number system merely by talking about the square root of $-1$ makes students think that complex numbers don't "exist" or are not important because it addresses a seemingly trivial problem.

I like this alot, its only deficiency (if there is one!) being there are not many pictures. :)
–
user641Dec 5 '12 at 17:19

@GregL I'm confused by the point you are making.
–
JebruhoDec 5 '12 at 17:32

3

You might get some nice pictures by linking this 'strange' new way of multiplying pairs of 2-vectors to the geometric interpretation of complex multiplication as a rotation and a dilation/contraction. For example, draw some points, lines, contour plots in the $x-y$ plane and then look at the effect under (say) multiplication by a constant $(x,y)\to(x,y)*(a,b)=(ax-by,bx+ay)$.
–
user12477Dec 5 '12 at 17:47

2

Is the ! a factorial, or are you really excited to explain math? :)
–
alexy13Dec 6 '12 at 1:07